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Question
Physics
Standing at the edge of a cliff 32.5m high, you do a ball. Later, you throw a second ball downward with an initial speed of 11m/s. Calculate the velocity if each of the balls when it hits the ground
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Best Answer

Answer: The displacement of the ball

ℎ =

2 −

2

2

So, final speed of the first ball

1 = √2ℎ = √2 × 9.8 × 32.5 = 25.24

m

s

The change of it speed

∆1 = 25.24 − 0 = 25.34

m

s

The final speed of the second ball

2 = √

2 + 2ℎ = 27.53

m

s

The change of the second ball speed

∆2 = 27.53 − 11 = 16.53

m

s

Thus

∆1 > ∆2

Answer: ∆1 > ∆2

Explanation:

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